Q22 Ex 3.7. Thomas Calculus Twelfth Edition Page 153 | University of Cambridge - Edubirdie

University:
University of Cambridge
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0580 IGCSE | Mathematics
Academic year:

2023
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454

Pages:

2
Author:

Shahbaz A.

Q 22 Ex 3.7 THOMAS’ CALCULUS Twelfth Edition Page 153 shahbaz ahmed August 2024 Q22 Use implicit differentiation to find dy dx y 2 − 2x = 1 − 2y y 2 − 2x = 1 − 2y d 2 d [y − 2x] = [1 − 2y] dx dx d 2 d d d [y ] − [2x] = [1] − [2y] dx dx dx dx d 2 dy dx dy [y ] −2 =0−2 dy dx dx dx dy dy − 2 = −2 dx dx dy dy 2y +2 =2 dx dx dy dy 2[y + ]=2 dx dx dy 2 dy + ]= [y dx dx 2 dy dy [y + ]=1 dx dx dy [y + 1] =1 dx dy 1 = dx y+1 2y dy dx = 1 y+1 dy 1 = dx y+1 1 and then d2 y dx2 dy = (y + 1)−1 dx d dy d [ ]= [(y + 1)−1 ] dx dx dx d dy d2 y = [(y + 1)−1 ] dx2 dy dx d2 y dy = (−1)[(y + 1)−2 ] dx2 dx 1 dy d2 y =− 2 2 dx (y + 1) dx Putting dy dx = 1 y+1 d2 y 1 1 2 =− [ ] dx2 (y + 1)2 y + 1 d2 y 1 =− 2 dx (y + 1)4 2

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Q22 Ex 3.7. Thomas Calculus Twelfth Edition Page 153

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